What is the sum from n=0 to infinity of 5/(4^n-3^n) ?

The sum from n=0 to infinity of 5/(4^n-3^n) is converges

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sum from n=0 to infinity of 5/(4^n-3^n)

n = 0 \sum \infty \frac{5}{4 ^{n} - 3 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{4 ^{n} - 3 ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 1 \sum \infty \frac{1}{4 ^{n} - 3 ^{n}}

Factor

4^{n} - 3^{n} : - (- 4^{n} + 3^{n})

= 5 \cdot n = 1 \sum \infty \frac{1}{- ( - 4 ^{n} + 3 ^{n} )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 (- n = 1 \sum \infty \frac{1}{- 4 ^{n} + 3 ^{n}})

Apply Series Ratio Test:

converges

= 5 (- converges)

= converges

n = 0 \sum \infty 7

n = 1 \sum \infty 8 n sin (\frac{7}{n})

n = 1 \sum \infty \frac{e ^{3 n}}{n !}

n = 5 \sum \infty (- 1)^{n} e^{- n}

n = 1 \sum \infty \frac{1 + ln ( n )}{n}

What is the sum from n=0 to infinity of 5/(4^n-3^n) ?
The sum from n=0 to infinity of 5/(4^n-3^n) is converges